Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.lang.*;

public class Rextester{
    public static void main(String[] args)throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int r = new Integer(st.nextToken());
        int c = new Integer(st.nextToken());
        int[][] matrix = new int[r][c];
        for(int i=0;i<r;i++){
            StringTokenizer st1 = new StringTokenizer(br.readLine());
            for(int j=0;j<c;j++){
                matrix[i][j]=new Integer(st1.nextToken());
            }
        }
        br.close();
        int[][] matrix2 = new int[r][c];
        StringBuffer sb = new StringBuffer();
        int moves=0;
        if(r<c){
            for(int i=0;i<r;i++){
                int min = matrix[i][0];
                for(int j=1;j<c;j++){
                    min=Math.min(min,matrix[i][j]);
                }
                for(int j=0;j<c;j++){
                    matrix2[i][j]=min;
                }
                for(int j=0;j<min;j++){
                    sb.append("row ").append(i+1).append("\n");
                }
                moves+=min;
            }
            for(int j=0;j<c;j++){
                int diff = matrix[0][j]-matrix2[0][j];
                for(int i=0;i<r;i++){
                    if(matrix[i][j]-matrix2[i][j]!=diff){
                        System.out.println("-1");
                        return;
                    }
                }
                moves+=diff;
                for(int i=0;i<r;i++){
                    matrix2[i][j]+=diff;
                }
                for(int i=0;i<diff;i++){
                    sb.append("col ").append(j+1).append("\n");;
                }
            }
        }
        else{
            for(int j=0;j<c;j++){
                int min = matrix[0][j];
                for(int i=1;i<r;i++){
                    min = Math.min(min,matrix[i][j]);
                }
                moves+=min;
                for(int i=0;i<min;i++){
                    sb.append("col ").append(j+1).append("\n");
                }
                for(int i=0;i<r;i++){
                    matrix2[i][j]=min;
                }
            }
            for(int i=0;i<r;i++){
                int diff = matrix[i][0]-matrix2[i][0];
                for(int j=0;j<c;j++){
                    if(diff!=matrix[i][j]-matrix2[i][j]){
                        System.out.println("-1");
                        return;
                    }
                }
                moves+=diff;
                for(int j=0;j<diff;j++){
                    sb.append("row ").append(i+1).append("\n");
                }
                for(int j=0;j<c;j++){
                    matrix2[i][j]+=diff;
                }
            }
        }
        for(int i=0;i<r;i++){
                for(int j=0;j<c;j++){
                    if(matrix[i][j]!=matrix2[i][j]){
                        System.out.println("-1");
                        return;
                    }
                }
            }
        System.out.println(moves+"\n"+sb);
    }
}